MST Topics in History of Mathematicsmath.fau.edu/yiu/MSTHM2017/17HM726B.pdf602 The ancient Chinese...

MST Topics inHistory of Mathematics

Euclid’s Elements,the Works of Archimedes,

and the Nine Chapters of Mathematical Art

Paul Yiu

Department of MathematicsFlorida Atlantic University

Chapters 17–21

Summer 2017

July 25, 2017

Contents

22 The ancient Chinese counting board 60122.1 The counting rods . . . . . . . . . . . . . . . . . . . . . . . . . . 60122.2 Addition and subtraction . . . . . . . . . . . . . . . . . . . . . . . 60222.3 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60222.4 Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60422.5 Ancient Chinese extraction of square root . . . . . . . . . . . .. . 606

22.5.1 Appendix: A paper-and-pencil algorithm for square roots . . 609

23 Fractions inFangtian 61123.1 Jiuzhang SuanshuI.1-4 . . . . . . . . . . . . . . . . . . . . . . . 61123.2 Reduction of fractions . . . . . . . . . . . . . . . . . . . . . . . . 612

23.2.1 Rule of reduction of fractionsL3 . . . . . . . . . . . . . . . 61223.3 Addition of fractions . . . . . . . . . . . . . . . . . . . . . . . . . 61223.4 Subtraction of fractions . . . . . . . . . . . . . . . . . . . . . . . 61323.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61423.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61523.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61623.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617

24 Shanggong: Volume of Solids 61924.1 Book XII of Euclid’s Elements . . . . . . . . . . . . . . . . . . . 61924.2 Book V of the Nine Chapters . . . . . . . . . . . . . . . . . . . . 62224.3 Hilbert’s Third Problem . . . . . . . . . . . . . . . . . . . . . . . 62424.4 The Nine Chapters . . . . . . . . . . . . . . . . . . . . . . . . . . 62524.5 Pyramids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626

25 Right Triangles in Zhoubi Suanjing 63125.1 The right triangle theorem . . . . . . . . . . . . . . . . . . . . . . 63125.2 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632

26 Right triangles: Chapter 9 of JZSS 637

27 Haidao Suanjing: LIU Hui’s Appendix to JZSS 659

Chapter 22

The ancient Chinese counting board

22.1 The counting rods

The numeration system using counting rods is a decimal positional system. A num-ber is represented by a collection of counting rods. This representation is simplifiedby first placing all the rods in the rightmost position in a (horizontal) row of squares,and replacing every group of ten rods by a single one in its immediate left neigh-bour. By applying this rule to every group of ten rods, beginning with the rightmostposition one eventually obtains the decimal representation of the number, consist-ing of a horizontal array of groups of counting rods, each notmore than nine, andpossibly empty. To avoid confusion, the rods in adjacent squares are arranged inperpendicular directions. If a square contains more than five rods, then five of theserods are replaced by one placed in a perpendicular direction.

1 2 3 4 5 6 7 8 9

vertical | || ||| |||| ||||| | | | ||| ||||

horizontal | | | |

For example, six thousand thirty seven is represented by

thousands hundreds tens units

| | |

with an empty space in the hundreds position.

602 The ancient Chinese counting board

22.2 Addition and subtraction

It is clear that addition and subtraction can be easily carried out simply by addingand removing rods. For subtraction, one often has to replacea rod in a square byten rods in its immediate right neighbour.

22.3 Multiplication

Problem I.1 ofSunzi Suanjinggives an example of how multiplication is carried outwith counting rods.1 It must be emphasized that although a succession of tables isshown here, the actual calculation is done in the same three lines, moving througha number of stages.

Nine nines are eighty one. Find the amount when this is multiplied byitself.Answer: 6561.Method: Set up the two positions: [upper and lower]. The upper 8calls the lower 8: eight eights are 64, so put down 6400 in the middleposition. The upper 8 calls the lower 1: one eight is 8, so put down 80in the middle position.Shift the lower numeral one place [to the right] and put away the 80 inthe upper position. The upper 1 calls the lower 8: one eight is8, so putdown 80 in the middle position. The upper 1calls the lower 1: one oneis 1, so put down 1 in the middle position. Remove the numerals in theupper and lower positions leaving 6561 in the middle position.

(1) The multiplication of large number depends on the multiplication of singledigit numbers. The standard Chinese begins withnine nines is 81down toone one is 1.

(2) Multplication is performed on a three-row counting board.

multiplicandproduct

multiplier

The unit of the multiplier is initially placed directly under the leftmost digit ofthe multiplicand.

1Translation by Lam and Ang, p.196.

22.3 Multiplication 603


multiplicand | |

product

multiplier | |


multiplicand | |

product | ||||

multiplier | |


multiplicand | |

product | |||| |

multiplier | |


multiplicand |

product | |||| |

multiplier | |


multiplicand |

product | |||| | | |

multiplier | |



multiplicand

product | ||||| | |

multiplier

22.4 Division

Problem I.2: If 6561 is divided among 9 persons, find how much eachgets.Answer: 729.Method: First set 6561 in the middle position to be the dividend (shi).Below it, set 9 persons to be the divisor (fa).Put down 700 in the upper position. The upper 7 calls the lower9:seven times nines are 63, so remove 6300 from the numeral in the mid-dle position.Shift the numeral in the lower position one place [to the right] and putdown 20 in the upper position. The upper 9 calls the lower 9: nne ninesare 81, so remove 81 from thenumeral in the middle position. There isnow no numeral in the middle position. Put away the numeral inthelower position. The result in the upper position is what eachpersongets.

quotientdividend

divisor

(1) The divisor is initially placed in the bottom line that itcan be subtracted bythe number “above it” in the dividend.

(2) The quotient is calledshangin Chinese, meaning “guess and see to figureout”. In each case, it amounts to finding the largest number which multiplied to thedivisor does not exceed the dividend. This can be done by repeatedly subtractingthe divisor from the dividend. The Chinese term for division is chu, removal.

22.4 Division 605


quotient

dividend | ||||| | |

divisor ||||


quotient | |

dividend | ||||| | |

divisor ||||


quotient | |

dividend || | |

divisor ||||


quotient | |

dividend || | |

divisor ||||


quotient | |

dividend | |

divisor ||||



quotient | | ||||

dividend | |

divisor ||||thousands hundreds tens units

quotient | | ||||

dividend

divisor ||||thousands hundreds tens units

quotient | | ||||

dividend

divisor

Lam considered the replacement of counting rods by the abacus (in the 16th cen-tury) a cause of decline of mathematics in China. “The use of the abacus to shortenthe time of calculation necessitated the rote learning of numerous mathematicalmethods. The rigorous step-by-step reasoning so essentialto the development ofmathematics is discarded, ad inevitably, mathematics declined.”

22.5 Ancient Chinese extraction of square root

In ancient China, arithmetic operations were performed using calculating rods ar-ranged on a counting board. The extraction of a square root iscarried out in a 4-rowarrangement, each column for one digit:

(i) The second row is labelledDividend. One begins by entering in this row thenumber whose square root is to be extracted.

(ii) The fourth row is labelledUnit. Beginning with a counting rod in the columnof the rightmost digit, one repeatedly shifts this 2 places to the left before it reachesbeyond the leftmost digit of thedividend.

(iii) The first row is labelledQuotient, in this row shall appear the square rootin question.

22.5 Ancient Chinese extraction of square root 607

(iv) The third row, labelledDivisor, is auxiliary in the computation.We illustrate this with Problem I.16 ofJiuzhang Suanshuon the extraction of

the square root of 3,972,150,625.

Quotient3 9 7 2 1 5 0 6 2 5 Dividend

Divisor1 Unit

The computation begins with finding the largest number whosesquare does notexceed the number with unit indicated in the bottom row. In this case, the largestsquare not exceeding 39 is62 = 36. Here one enters 6 in the first row,2 and subtract36 from the second row.

6 Quotient3 7 2 1 5 0 6 2 5 Dividend

Divisor1 Unit

Multiply the quotient by 2, enter it in the third row (asdivisor), shifting oneplace to the right. Shift also the unit rod in the bottom row two places to the right.

6 Quotient3 7 2 1 5 0 6 2 5 Dividend1 2 Divisor

1 Unit

In the column of the unit rod, put thelargestnumber which, when multiplied tothe numberwith the same units digit in the third row, gives a product not exceedingthe number in the second row. In the present case, this numberis 3. Subtract theproduct from the second row.

6 3 Quotient3 1 5 0 6 2 5 Dividend

1 2 3 Divisor1 Unit

Double the number in the first row, enter it in the third row,3 shifting one placeto the right. Shift the unit rod in the bottom row two places tothe right.

2The square root in question being a 5-digit number, the leftmost digit 6 should be entered in the5th column from the right. There is, however, definite advantage in aligning it with the unit rod inthe bottom row.

3This has the same effect asdoublingthe unit digits of the divisor in the third row.


6 3 Quotient3 1 5 0 6 2 5 Dividend

1 2 6 Divisor1 Unit

Repeat the same procedure till the unit rod appears at the end. The numberappearing in the first row would be the square root.

In the present example, the next number in the first row is 0. Thus, one simplyshifts the divisor one place to the right, and the unit rod twoplaces to the right.

6 3 0 Quotient3 1 5 0 6 2 5 Dividend1 2 6 0 Divisor

1 Unit

6 3 0 2 Quotient6 3 0 2 2 5 Dividend

1 2 6 0 2 Divisor1 Unit

6 3 0 2 Quotient6 3 0 2 2 5 Dividend1 2 6 0 4 Divisor

1 Unit

The next number should be 5. Since5 × 126045 = 630225, this leaves thesecond row blank, and the calculation terminates, giving 63025 for the square root.

6 3 0 2 5 QuotientDividend

1 2 6 0 4 5 Divisor1 Unit

If the dividend appearing in the last step is a nonzero numberr, the square rootin question is not “exact”. In this case, the ancient Chinese adopted one of thefollowing options.

(i) Round off the square root with the fractionr2q+1

or r2q

, q being the quotientin the first row. In other words,

√

q2 + r ≈ q +r

2q + 1or q +

r

2q.

(ii) Continue the calculation “beyond the decimal point” by treating a unit as 10subunits (fen), 100 sub-subunits (li ), 1000 sub-sub-subunits (hou), etc.

22.5 Ancient Chinese extraction of square root 609

Exercise

Find the square roots of the following numbers (i) 55225, (ii) 25281, (iii) 71824,(iv) 564, 7521

4.

A diagrammatic explanation of the procedure of extraction of square roots

In Jiuzhang Suanshu, all square roots are exact. In explaining the extraction ofsquare root, the text says

dang yi mian ming zhi.

Earlier commentators interpreted this as rounding off the square root with thefractions r

2q+1or r

2q. LI Jimin, however, explains thatmianrefers to the square root

itself, so that what the text means is thatQ = · · ·+ rQ

.

22.5.1 Appendix: A paper-and-pencil algorithm for square roots

To find the square root of an integera, divide the digits ofa into blocks of two digitsbeginning with the right hand side. We represent this as

a1a2 · · · an

where eachak, k = 2, . . . , n is a 2-digit number, anda0 has either 1 or 2 digits.

(1) Setb1 := a1, and letq1 be the largest integer such thatr1 := a1 − q21 ≥ 0.SetQ1 := q1.

(2) Supposebn, rn andQn have been defined. Form

bn+1 := 100rn + an+1.

Find thelargest integerqn+1 such that

rn+1 := bn+1 − (20Qn + qn+1)qn+1 ≥ 0.

SetQn+1 := 10Qn + qn+1.


2 5 5 8 6 6 8 5 06 5 4 6 7 8 4 5 3 6 1 2 3 4 5 6 74

45 2 5 42 2 5

505 2 9 6 72 5 2 5

5108 4 4 2 8 44 0 8 6 4

51166 3 4 2 0 5 33 0 6 9 9 6

511726 3 5 0 5 7 6 13 0 7 0 3 5 6

5117328 4 3 5 4 0 5 2 34 0 9 3 8 6 2 4

51173365 2 6 0 1 8 9 9 4 52 5 5 8 6 6 8 2 5

511733700 4 3 2 3 1 2 0 6 7

4 3 2 3 1 2 0 6 7

Chapter 23

Fractions in Fangtian

23.1 Jiuzhang SuanshuI.1-4

I.1 There is a field 15bu wide and 16bu long.Question: How much is (the area of) the field?Answer: 1mu.

I.2 There also is a field 12bu wide and 14bu long.Question: How much is (the area of) the field?Answer: 168 [square]bu.

Rectangular field rule:

Multiply the width and the length to obtain the (number of) squarebu. L1 Dividethis by 240, the standard ofmu. This is (the area in) number ofmu. One hundredmuis oneqıng.

L1: This product is the area. Every product of a width and a length is calledmı.

I.3 There is a field 1l ı wide and 1l ı long. How much is (the area of) the field?Answer: 3qıng75mu.

I.4 There also is a field 12bu wide and 14bu long. How much is (the area of)the field?Answer: 168 [square]bu.

Rule of field in l ı

Multiply the numbers ofl ı in width and length to obtain (the number of) squarel ı;multiply this by 375. This is (the area in) number ofmu.L2

L2: Note that in this method the numbers ofl ı in width and length multiply togive the number of squarel ı. There are 3qıng 75 mu in a squarel ı, so multiplyingby this number, one obtains the number inmu.

612 Fractions in Fangtian

23.2 Reduction of fractions

I.5 There are 12 parts of 18. How much are these reduced to?Answer: 2 parts of 3.

I.6 There also are 49 parts of 91. How much are these reduced to?Answer: 7 parts of 13.

23.2.1 Rule of reduction of fractionsL3

If (both quantities are) halvable, half them. If not (both) halvable, place the denom-inator and numerator side by side. Keep on subtracting the lesser from the more, tofind (two) equal numbers. Now, reduce (the denominator and the numerator) by the(value of the) equal numbers.L4

L3: As to reduction of fractions, note that the quantities of things may not al-ways be whole, and it is necessary to speak of fractions. Fractions are difficult touse if they are too complicated. Suppose there are 2 parts of 4. In a more compli-cated expression, these are the same as 4 parts of 8. However,in reduced for, thisis 1 part of 2. The expressions might be different, but as numbers, they serve thesame end. In comparing the divisor with the dividend, one often finds unevenness.Therefore, the practitioner must first learn to deal with fractions.

L4: To reduce by thedeng shu is to divide by it. Those (numbers) involved inthe subtractions being all multiples of thedeng shu, we reduce (the given quantities)by it.

23.3 Addition of fractions

I.7 There are one parts of three, two parts of five. How much do we get by addingthem?Answer: eleven parts of fifteen.

I.8 There are also two parts of three, four parts of seven, andfive parts of nine.How much do we get by adding them?Answer: get one and fifty parts of sixty three.

I.9 There are also two parts of two, two parts of three, three parts of four, andfour parts of five. How much do we get by adding them?Answer: get two and forty three parts of sixty.

Rule of addition of fractions

Let the denominators cross multiply (hucheng) (each) numerator; combine the re-sults as dividend. Multiply the denominators together for the divisor.L5 Compare

23.4 Subtraction of fractions 613

the dividend to the divisor as unit. If there is any remainder, make it the numeratorof a fraction with the divisor as denominator.L6 When all denominators are thesame, directly add up (the numerators).

L5: Let the denominators cross multiply (hucheng) (each) numerator, (A frac-tion) in its reduced form has a coarse subdivision (of its denominator), whereaswith a complicated expression, such a subdivision is fine. Fine or coarse, they rep-resent the same fraction. Given a number of fractions, intricate and mixed (in thattheir denominators are different), one cannot combine themwithout finely subdi-viding (their denominators). Multiply (both denominator and numerator of eachfraction by the same number) tosan the fractions, so as tocommunicatethem. Witha common denominator, (the numerators) can be combined together. Letting thedenominators cross multiply (hucheng) (each) numerator is called homogenize (qı). Multiplying the denominators together is called uniformize (tong). To uniformize(Tong) is to multiply the denominator and the numerator of each fraction by thesame number, so as to make the fractions uniformize (tong), i. e. sharing a com-mon denominator. To homogenize (Qı) is to align the numerator and the denomi-nator; it is essential not to alter the value (of the fraction). Methods are classifiedinto types; things are divided into groups. Numbers of the same kind are not far;whereas those of different kinds are not near. For those withthe same denominator,even their numerators are far in that they occupy different places (on a calculatingboard), these can be added together. (On the other hand), forthose with differentdenominators, even if their numerators are so near as to be inthe same column, theyare still at odd with one another. The methods of homogenize (qı) and uniformize(tong) are very important indeed! The (given) fractions may be in disaccord; but(by such methods) we can bring them into harmony. This is likeequipping with anivory cone for untying knots, always managing (the fractions) orderly. Multiplyingto san fractions, dividing to reduce them, homogenize (qı) and uniformize (tong) tomake themtong. Is this not the guiding principle of calculation? As to the method,we may letlu be the product of all but one denominators, and multiply thislu to thenumerator to get homogenize (qı).

L6: Now to find the dividend, first homogenize (qı) the numerators an make thedenominators common. Compare (this dividend) with the divisor as unit. Reduceany remainder (with the divisor) by thedeng shu. That is it. This is to say thecommon divisor as denominator, and the remainder from the dividend as numerator.All (additions of fractions) follow these examples.

23.4 Subtraction of fractions

I.10 There are 8 parts of 9. Subtract from it 1 part of 5. How much is the remainder?Answer: 31 parts of 45.


I.11 There also are 3 parts of 4. Subtract from it 1 p of 3. How much is theremainder?Answer: 5 parts of 12.

Rule for subtraction of fractions

Let the denominators cross multiply (hucheng) the numerators. Subtract the smaller(product) from the greater,L7 (and take) the remainder as dividend. Multiply thedenominators to form the divisor. (Compare) the dividend to the divisor as unit.

I.12 There are 5 parts of 8, and 16 parts of 25. Which is more, by how much?Answer: 16 parts of 25 is more, by 3 parts of 200.

I.13 There also are 8 parts of 9, and 6 parts of 7. Which is more, by how much?Answer: 8 parts of 9 is more, by 2 parts of 63.

I.14 There are 8 parts of 21, and 17 parts of 50. Which is more, byhow much?Answer: 8 parts of 21 is more, by 13 parts of 1050.

Rule for comparison of fractions

Let the denominators cross multiply (hucheng) the numerators. Subtract the smaller(product) from the greater,L7 (and take) the remainder as dividend. Multiply thedenominators to form the divisor. (Compare) the dividend to the divisor as unit.That is the difference

L7: (To let) the denominators cross multiply (hucheng) the denominator is tohomogenize (qı) the numerators. (We may) subtract the smaller from the greatersince the numerators are homogeneous (qı). To multiply the denominators togetheras the divisor is to make a common denominator. Since the denominator and thenumerator are aligned, (we may) compare (the difference) tothe denominator asunit. QED.

23.5

I.15 There are one part of three, two parts of three, and threeparts of four. Howmuch should be subtracted from the more to benefit the fewer tomake themthe same level?Answer: subtract two from three parts of four, one from two parts of three,combine them to benefit one part of three; then they are all level at seven partsof twelve.

I.16 There are one part of two, two parts of three, and three parts of four. Howmuch should be subtracted from the more to benefit the fewer, to make them

23.6 615

the same level?Answer: subtract one from two parts of three, four from threeparts of four,combine them to benefit one part of two; then they are all levelat twenty threeparts of thirty six

Rule for averaging of fractions

Let the denominators cross multiply (hucheng) each numerator,L8. Combine (theproducts) to be thepengshı. Mutliply the denominators together to form the divi-sor.L9 Multiply each summand (of thepengshı) by the number of columns to formthe column dividends. Also multiply the divisor by the number of columns. Sub-tract thepengshı from the column dividends. Reduce the difference and take thesefor (the amounts) to be subtracted. Combine the amounts subtracted to benefit thesmallest one. Form a fraction with thepengshı as numerator and the division asdenominator. (Now), each fraction achieves the average.

L8: This is to align the numerators.L9: (Note on) multiplying the denominators together for the divisor: after align-

ing their numerators, now make their denominators common.L10: Here, we should have divided thepengshı by the number of columns. But

then, we would run into compound fractions. So, instead, we multiply the divisorby the number of columns.

23.6

I.17 There are seven persons dividing eightqian, one part of threeqian. Howmuch does each one get?Answer: each gets oneqian, four parts of twenty oneqian.

I.18 There also are three persons, one part of three persons,dividing sixqian, onepart of threeqian, and three parts of fourqian. How much does each one get?Answer: each gets twoqian, one part of eightqian.

Rule for division of fractions Method

The number of persons as divisor, the number ofqian as dividend, compare thedividend with the divisor as unit. Convert any mixed fractions into improper frac-tions.L11 For compound fractions, make he denominators (of the denominator andthe numerator) common.L12

L11: To let the denominators cross multiply (hucheng) the numerators is to alignthe numerators; to multiply the denominators is to make the denominators common.To uniformize (tong) with the denominators is to multiply the whole number by thedenominator and to bring (the product) into the numerator. Multiplying the whole


number (by the denominator) is tosan it into a j ıfen. Then thej ifenand the fractionpart aretong, and can be added up together. Every number used to multiply to boththe denominator and the numerator of a fraction is called1u. The (various)1u areso chosen to make the fractionstong. If there are mixed fractions, convert them intoimproper fractions. If the denominator and the numerator of) the fraction containcommon multiples, reduce them. Divide the divisor and the dividend by thedenyshu. This deny shu is thelu to multiply (to the denominator and the numerator) tomake the fraction (assume its current form). Therefore, tosan the fractions (in thedenominator and the numerator), we let both denominators multiply the divisor andthe dividend.

L12: Also, (when one of the numerator and the denominator is a whole num-ber), multiply the dividend by the denominator of the divisor, or the ’visor by thedenominator of the dividend, (as appropriately). This (last statement in the Method)is to say, if the divisor and the dividend are both mixed fractions, then we multiplyeach denominator to the whole number and bring it into the numerator, and thenlet he denominators cross multiply (hucheng) (the numerators) in the top and thebottom (of the compound fraction).

23.7

I.19 There is a field 7 parts of 9bu wide, 9 parts of 11bu long. How much is (thearea of) the field?Answer: 7 parts of 11bu.

1.20 There also is a field 4 parts of 5bu wide, 5 parts of 9bu long. How much is(the area of) the field?Answer: 4 parts of 9bu.

Rule for multiplication of fractions

Multiply the denominators together for the divisor; multiply the numerators to-gether for the dividend. Compare the dividend to the divisor as unit.L13

L13: Whenever (the dividend) does not measure up to the divisor, (one obtains aproper fraction with) denominator. and numerator (only). If there is (second) frac-tion which, when multiplied to it, makes the dividend exceedthe divisor, then thereresults in a whole number as well. Also, as the numerators aremultiplied together,the denominators must be effected in division,i.e., comparing the dividend to thedivisor as unit. Since multiplying the numerators togetherrequires at the same timethat the denominators be effected into division, we multiply the denominators to-gether and perform division subsequently. In speaking about (areas of rectangular)fields, (one encounters two analogous quantities) width andlength, and it is diffi-

23.8 617

cult to illustrate (with such the connection between this and the previous method).Consider, therefore, the problem:

20 horses are worth 12jin of gold. Now, 20 horses are sold and theprofit is shared among 35 persons. How much does each one get?Answer: 12 parts of 35jin.

This follows from the division of fractions method, with 12jin of gold as divi-dend and 35 persons as divisor. Let us now reformulate the problem:

5 horses are worth 3jin of gold. Now 4 horses are sold and the profit isshared among 7 persons. How much does each one get?Answer: 12 parts of 35jin of gold.

This follows by aligning the numbers of (jin) gold and persons, and then therule for division of fractions, as in the previous problem. Now, to multiply the nu-merators for the dividend is like aligning the (number ofjin) gold, and multiplyingthe denominators for the divisor is like aligning the (number of) persons. The com-mon denominator 20 has no relevance in the calculation. One simply wants to alignthe (numbers ofjin of gold and persons). Also, for 5 horses worth 3jin of gold,these number are all whole. In terms of fractions, a horse is worth 3 part of 5jinof gold. (Since) 7 persons sell 4 horses, (one may regard) each person selling 4parts of 7 horses. (The answer is now obtained from) the numerator (the dividendin the division above) and turning upside down (the divisor,the fraction obtainedby considering) “persons” (as above), by multiplying together. The expressions aredifferent, but all three methods lead to the same answer.

23.8

I.21 There is a field 3bu and 1 part of 3bu wide, 5bu and parts of 5bu long. Howmuch is (the area of) the field?Answer: 18bu.

I.22 There also is a field 7bu and 3 parts of 4bu wide, 15 b’ and 5 parts of 9bulong. How much is (the area of) the field?Answer: 120bu and 5 parts of 9 bu.

I.23 There also is a field 18bu and 5 parts of 7bu wide, 2bu and 6 parts of 11bulong. How much is (the area of) the field?Answer: 1 mu 200bu and 7 parts of 11 bu.


Rule for most general field

Multiply each denominator t the whole number, and add to the numerator.L14 Multi-ply these (sums) for the dividend, and multiply the denominators for the divisor.L15

Compare the dividend to the divisor as unit.L16

L14: Multiply each denominator to the whole number and ad to the numerator.This is to tong the whole number ofbu and bring into the numerator. Then both thedenominator and the numerator are whole numbers.

L15: (This is) like the multiplication of fractions (above).L16: In this Method, both width and length are mixed f actions. Each (of these)

should be converted into an improper fraction. As the denominators have be broughtinto (their own numerators), they need to be brought out again (after the multipli-cation). We therefore multiply the denominators for the divisor and carry out thedivision subsequently.

Chapter 24

Shanggong: Volume of Solids

24.1 Book XII of Euclid’s Elements

Eucl. XI. Definition 12

A pyramid is a solid figure, contained by planes, which is constructed from oneplane to one point.

Notation:(V,ABC . . .K).

Eucl. XI. Definition 13

A prism is a solid figure contained by planes two of which, namely those which areopposite, are equal, similar, and parallel, while the rest are parallelograms.

A right triangular prism is a right triangular prism. Its two bases are congruentright trianglesABC andA′B′C ′ in which the linesAA′, BB′, CC ′ are perpendic-

ular to the planes ofABC andA′B′C ′. We shall write this as

(

ABCA′B′C ′

)

.

Eucl. XII.3

Any pyramid with a triangular base is divided into two pyramids equal and similarto one another, similar to the whole, and having triangular bases, and into two equalprisms, and the two prisms are greater than half of the whole pyramid.

Outline proof

(Heath) We will denote a pyramid with vertexD and baseABC by D(ABC)or D − ABC and the triangular prism with trianglesGCF , HLK for bases by(GCF,HLK). The following are the steps of the proof.

620 Shanggong: Volume of Solids

A B

D

H K

C

G

F

E

L

Figure 24.1:

I. To prove pyramidH(AEG) equal and similar to pyramidD(HKL). Sincesides of△DAB are bisected atH, E, K,

HE//DB, and HK//AB.

HenceHK = EB = EA,

andHE = KB = DK.

Therefore (1)△HAE and△DHK are equal and similar.Similarly (2)△HAG and△DHL are equal and similar.Again,LH, HK are respectively parallel toGA, AE in a different plane; there-

fore∠GAE = ∠LHK.AndLH, HK are respectively equal toGA, AE.Therefore (3)△GAE and△LHK are equal and similar.Similarly (4)△HGE and△DLK are equal and similar.Therefore1 the pyramidsH(AEG) andD(HKL) are equal and similar.II. To prove the pyramidD(HKL) similar to the pyramidD(ABC). 2

(1)△DHK and△DAB are equiangular and therefore similar.Similarly (2)△DLH and△DCA are similar, as also(3)△DLK and△DCB.Again,BA, AC are respectively parallel toKH, HL in a different plane; there-

fore∠BAC = ∠KHL. AndBA : AC = KH : HL.Therefore (4)△BAC and△KHL are similar.Consequently the pyramidD(ABC) is similar to the pyramidD(HKL), and

therefore also to the pyramidH(AEG).III. To prove prism(GCF,HLK) equal to prism(HGE,KFB).

1Eucl. XI, Definition 10: Equal and similar solid figures are those contained by similar planesequal in multitude and in magnitude.

2Eucl. XI, Definition 9: Similar solid figures are those contained bysimilar planes equal inmultitude.

24.1 Book XII of Euclid’s Elements 621

The prisms may be regarded as having the same height (the distance betweenthe planesHKL, ABC) and having for bases

(1)△CGF and(2) the parallelogramEBFG, which is the double of△CGF .Therefore, by XI.39,3 the prisms are equal.IV. To prove the prisms greater than the small pyramids. Prism (HGE,KFB)

is clearly greater than pyramidK(EFB) and therefore greater than pyramidH(AEG).Therefore each of the prisms is greater than each of the smallpyramids; and the

sum of the two prisms is greater than the sum of the two small pyramids, which,with the two prisms, make up the whole pyramid.

Eucl. XII.4

If there are two pyramids of the same height with triangular bases, and each ofthem is divided into two pyramids equal and similar to one another and similar tothe whole, and into two equal prisms, then the base of the one pyramid is to thebase of the other pyramid as all the prisms in the one pyramid are to all the prisms,being equal in multitude, in the other pyramid.

Eucl. XII.5

Pyramids of the same height with triangular bases are to one another as their bases.

Eucl. XII.7

Any prism with a triangular base is divided into three pyramids equal to one anotherwith triangular bases.

Porism

From this it is manifest that any pyramid is a third part of theprism which has thesame base with it and equal height.

(i) C(ABD) = (C,DEB) = (D,EBC) by XII.5.(ii) (D,EBC) = (D,ECF ).Thus, (C,ABD) = (D,EBC) = (D,ECF ). Since these three pyramids

together make up the prism, each of them is one third of the prism.

3If there be two prisms of equal height, and one have a parallelogram as base and the other atriangle, and if the parallelogram be double of the triangle, the prisms will be equal.


B A

C

E D

F

Figure 24.2:

24.2 Book V of the Nine Chapters

Book V of the Nine Chapters, with the title Shanggong(Construction consulta-tions), is on the mensuration of solids. The first 9 problems are on the volume ofprisms of uniform section.

V.10

Now given a frustum of a pyramid, with a lower section 5zhangsquare, an uppersection 4zhangsquare and an altitude of 5zhang. Question: What is the volume ?

The rule for a frustum of a pyramid

Multiply the side of the upper square by that of the lower. Square each of them andadd [the products]. Multiply [the sum] by the altitude and divide by 3.

V.11

Now given a frustum of a cone with a lower circumference of 3zhang, an uppercircumference of 2zhangand an altitude of 1zhang. Question: What is the volume?


Multiply the side of the upper circumference by that of the lower. Square each ofthem and add [the products]. Multiply [the sum] by the altitude and divide by 36.

V.12

Now given a pyramid, with a base 2zhang7 chi square, and an altitude of 2zhang9 chi. Question: What is the volume ?

24.2 Book V of the Nine Chapters 623

The rule for a pyramid

Square the side, multiply by the altitude, divide by 3.

V.13

Now given a circular cone with a base circumference of 3zhang5chiand an altitudeof 5 zhang1 chi. Question: What is the volume ?

The rule for a circular cone

Square the base circumference, multiply by the altitude, divide by 36.

V.14

Now given a right triangular prism with a lower breadth of 2zhang, a length of 18zhangand an altitude of 2zhang5 chi. Question: What is the volume ?

The rule for a right triangular prism

Multiply the breadth by the length, then multiply by the altitude, divide by 2.

V.15

Now given a roof-corner with a breadth of 5chi, a length of 7chi and an altitude of8 chi. Question: What is the volume ?

LIU Hui’s commentary on the volume of a roof-corner

Cut a cuboid into two right triangular prisms, and dissect a right triangular prismalong a diagonal to give a roof-corner and a turle-shoulder.

Claim: the roof-corner to the turtle shoulder is 2 parts to 1 part. These are fixedrates.

[When the cuboid is a cube], two turtle-shoulder constitute one roof-corner, and3 roof-corners constitute 1 cube. So divide by 3. This is clear when their spatialrelationship is verified by standard blocks. For a general cuboid, however, the 6turtle-shoulders are not congruent. But the three sides are equal correspondingly,and their volumes are equal. The turtle-shoulders are different in shape, and theroof-corners of different forms, making it difficult to compare their volumes. Howso ?

Consider a right triangular prism with a top and right faces, and half of thefront and rear faces of a cuboid. Cut it into an inner roof-corner and an outer turtle


shoulder. [This means that the rear bottom vertex is joined to the vertices of the topface to the roof-corner, and the turtle-shoulder has the front face and the right faceof the prism.

Color the roof-corner red and the turtle-shoulder black. Reassemble them to-gether to form a right triangular prism, and cut along the mid-sections in all threedirections.

From the red roof-corner we have one cuboid (rear right hand corner top), tworight triangular prisms (front right top and rear left top) and two roof corners (rearright bottom). The two right triangular prisms can be put together to form a cuboid.So we have

one large roof-corner = 2 small cuboids + 2 small roof-corners.

From the black turtle-shoulder we have two small turtle-shoulders (top left frontand bottom right rear, of the same shape) and two right triangular prisms (top front,and top bottom). These two right triangular prisms are not ofthe same shape. Butclearly together they have the same volume of a small cuboid.

one large turtle-shoulder = 1 small cuboid + 2 small turtle-shoulders.

Cubes from blocks different from the original forms have a rate of 3, while acube with the original form of the blocks has a rate of 1. Even with right triangularprisms of different breadth, length and altitude the same conclusion holds. If theremaining blocks [turtle-shoulders and roof-corners] arestill in the ratio of onepart to two parts, the rates of 1 and 2 are fixed. Therefore the argument is notfalse. To exhaust the calculation, halve the remaining breadth, length and altituderespectively, and an additional three quarters can thus be determined. The smallerthe halves, the finer the remainder. Extreme fineness means infinitesimal, which isformless. In that case. how can one have a remainder?

24.3 Hilbert’s Third Problem

The equality of the volumes of two tetrahedra of equal bases andequal altitudes

In two letters to Gerling, Gauss4 expresses his regret that certain theorems of solidgeometry depend upon the method of exhaustion,i.e., in modem phraseology, uponthe axiom of continuity (or upon the axiom of Archimedes). Gauss mentions in par-ticular the theorem of Euclid, that triangular pyramids of equal altitudes are to eachother as their bases. Now the analogous problem in the plane has been solved: Ger-ling also succeeded in proving the equality of volume of symmetrical polyhedra byby dividing them into congruent parts. Nevertheless, it seems to me probable that

4Werke, vol. 8, pp.241 and 244.

24.4 The Nine Chapters 625

a general proof of this kind for the theorem of Euclid just mentioned is impossible,and it should be our task to give a rigorous proof of its impossibility. This would beobtained, as soon as we succeeded inspecifying two tetrahedra of equal basesand equal altitudes which can in no way he split up into congruent tetrahe-dra, and which cannot be combined with congruent tetrahedra to form twopolyhedra which themselves could he split up into congruent tetrahedra. 5

24.4 The Nine Chapters

Book V of the Nine Chapters, with the title Shanggong(Construction consulta-tions), is on the mensuration of solids. The first 9 problems are on the volume ofprisms of uniform section.

V.10

Now given a frustum of a pyramid, with a lower section 5zhangsquare, an uppersection 4zhangsquare and an altitude of 5zhang. Question: What is the volume ?


Multiply the side of the upper square by that of the lower. Square each of them andadd [the products]. Multiply [the sum] by the altitude and divide by 3.

LIU Hui’s commentary

In this Chapter, theqiandu (right triangular prism) andyangma blocks combineto form a cuboid. So mathematicians have designed three kinds of blocks, calledqi, to model different solids. Suppose a frustum of a pyramid has an upper section1 chi square, a lower section 3chi square and an altitude of 1chi. The blocks usedare:1 cuboid in the center,4 qiandu at the sides, and4 yangma at the corners.

“Multiply the side of the upper square by that of the lower” gives 3 [square]chi; “multiply by the altitude” gives a volume of 3 [cubic]chi. This gives 1 centralcuboid and 4 lateralqiandu. The square of the lower side is 9 [square]chi; multiplyit the altitude, giving 9 [cubic]chi. This gives 1 central cuboid, 2qiandu on foursides, and threeyangma at four corners. The square of the upper side multiplied

5Since this was written Herr Dehn has succeeded in proving this impossibility. See his note[hUber raumgleiche Polyheder in Nach. d. K. Geselsch. d Wiss. zu Gottingen, 1900, and a papersoon to appear in Math. Annalen [vol. 55, pp. 465–178].


B A

C

E D

F

Figure 24.3:

by ltIealtitude gives 1 [cubic]chi, which is another central cuboid. Now the sumof the three kinds of blocks is three times as many as in the actual frustum of apyramid. Therefore, dividing it by 3 gives the volume in [cubic] chi. The totalnumber of blocks used is 27: 3 cuboids, 12qiandu and 12yangma, with whichthree frusta of a pyramid can be constructed, i.e. equal to 13cuboids. This is theverification. Alternatively, square the difference between the sides of the squaresections, multiply the altitude, divide by 3. This gives 4yangma. Multiply [thesides of] the upper and lower squares and then the altitude. This gives the centralcuboid and the 4 lateralqiandu. Add them, giving the volume of the frustum of apyramid.

V.11

Now given a frustum of a cone with a lower circumference of 3zhang, an uppercircumference of 2zhangand an altitude of 1zhang. Question: What is the volume?


Multiply the side of the upper circumference by that of the lower. Square each ofthem and add [the products]. Multiply [the sum] by the altitude and divide by 36.

24.5 Pyramids

V.12

Now given a pyramid, with a base 2zhang7 chi square, and an altitude of 2zhang9 chi. Question: What is the volume ?

24.5 Pyramids 627

The rule for a pyramid

Square the side, multiply by the altitude, divide by 3.

V.13

Now given a circular cone with a base circumference of 3zhang5chiand an altitudeof 5 zhang1 chi. Question: What is the volume ?

The rule for a circular cone

Square the base circumference, multiply by the altitude, divide by 36.

V.14

Now given aqiandu (right triangular prism) with a lower breadth of 2zhang, alength of 18zhangand an altitude of 2zhang5 chi. Question: What is the volume?

The rule for a right triangular prism


V.15

Now given ayangma, with a breadth of 5chi, a length of 7chi and an altitude of 8chi. Question: What is the volume ?

The rule for a yangma

Multiply the breadth by the length, then multiply by the altitude, divide by 3.Cut a cuboid into two right triangular prisms, and dissect a right triangular prism

along a diagonal to give a roof-corner and a turle-shoulder.Claim: the roof-corner to the turtle shoulder is 2 parts to 1 part. These are fixed

rates.[When the cuboid is a cube], two turtle-shoulder constitute one roof-corner, and

3 roof-corners constitute 1 cube. So divide by 3. This is clear when their spatialrelationship is verified by standard blocks.

For a general cuboid, however, the 6 turtle-shoulders are not congruent. Butthe three sides are equal correspondingly, and their volumes are equal. The turtle-shoulders are different in shape, and the roof-corners of different forms, making itdifficult to compare their volumes. How so ?


Consider a right triangular prism with a top and right faces, and half of thefront and rear faces of a cuboid. Cut it into an inner roof-corner and an outer turtleshoulder. [This means that the rear bottom vertex is joined to the vertices of the topface to the roof-corner, and the turtle-shoulder has the front face and the right faceof the prism.

Color the roof-corner red and the turtle-shoulder black.Reassemble them together to form a right triangular prism, and cut along the

mid-sections in all three directions.From the red roof-corner we have one cuboid (rear right hand corner top), two

right triangular prisms (front right top and rear left top) and two roof corners (rearright bottom). The two right triangular prisms can be put together to form a cuboid.So we have

one large roof-corner = 2 small cuboids + 2 small roof-corners.From the black turtle-shoulder we have two small turtle-shoulders (top left front

and bottom right rear, of the same shape) and two right triangular prisms (top front,and top bottom). These two right triangular prisms are not ofthe same shape. Butclearly together they have the same volume of a small cuboid.

one large turtle-shoulder = 1 small cuboid + 2 small turtle-shoulders.Cubes from blocks different from the original forms have a rate of 3, while a

cube with the original form of the blocks has a rate of 1. Even with right triangularprisms of different breadth, length and altitude the same conclusion holds. If theremaining blocks [turtle-shoulders and roof-corners] arestill in the ratio of onepart to two parts, the rates of 1 and 2 are fixed. Therefore the argument is notfalse. To exhaust the calculation, halve the remaining breadth, length and altituderespectively, and an additional three quarters can thus be determined. The smallerthe halves, the finer the remainder. Extreme fineness means infinitesimal, which isformless. In that case. how can one have a remainder?

V.16

Now given abie’nao (turtle bone), with a lower breadth of 5chi, no length; anupper length of 4chi, no breadth; and an altitude of 7chi. Question: What is thevolume ?

The rule for a bie’nao


V.17

Now given axianchu, with a lower breadth of 6chi, an upper breadth of 1zhang,a depth of 8chi. Question: What is the volume ?

24.5 Pyramids 629

The rule for a xianchu

Add the three breadths, multiply the sum by the length, then multiply by the depth,divide by 6.

V.18

Now given achumeng, with a lower breadth of 3zhang, a length of 4zhang, anupper length of 2zhang, no breadth, an altitude of 1zhang. Question: What is thevolume ?

The rule for a chumeng

Double the lower length, add it to the upper length, multiplythe sum by the breadth,then multiply by the altitude, divide by 6.

V.19

Now given achutong, with a lower breadth of 2zhang, a length of 3zhang, anupper breadth of 3zhang, a length of 4zhang; an altitude of 3zhang. Question:What is the volume ?

V.20

Now given aquchi, with an upper inner circumference of 2zhang, an outer cir-cumference of 4zhang, a breadth 1zhang; lower inner circumference of 1zhang4chi, an outer circumference of 2zhang4 chi, a breadth of 5chi; a depth of 1zhang.Question: What is the volume ?

V.21

Now given apanchi, with an upper breadth of 6zhang, a length of 8zhang; a lowerbreadth of 4zhang, a length of 6zhang; a depth of 2zhang. Question: What is thevolume ?

V.22

Now given aminggu, with an upper breadth of 2zhang, a length of 7zhang; alower breadth of 8chi, a length of 4zhang; a depth of 6zhang5 chi. Question:What is the volume ?


Exercise

V.1 The first 9 problems of Book V of theNine Chapters

V.1 Now given an excavation of 10,000 [cubic]chiof mud. Question: Howmuch rammed earth or loam comes from it?

V.2 given a city wall with a lower breadth of 4zhang, and upper breadth of 2zhang, an altitude of 5zhangand a length of 126zhang5 chi. Question:What is the volume ?

V.3 Now given a wall with a lower breadth of 4chi, and upper breadth of 2chi, an altitude of 1zhang2 chi and a length of 22zhang5 chi 8 cun.Question: What is the volume ?

V.4 Now given a dyke with a lower breadth of 2zhang, and upper breadth of8 chi, an altitude of 4chi 2 chi and a length of 12zhang7 chi. Question:What is the volume ?

Each laborer’s standard winter quota for earthworks is 444 [cubic] chi.Question: How many laborers are required?

V.5 Now given a trench with an upper breadth of 1zhang5 chi, a lowerbreadth of 1zhang, a depth of 5chi and a length of 7zhang. Question:What is the volume?

V.6 Now given a moat with an upper breadth of 1zhang6 chi 3 cun, a lowerbreadth of 1zhang, a depth of 6chi 3 cunand a length of 13zhang2 chi1 cun. Question: What is the volume?

V.7 Now given a canal with an upper breadth of 1zhang8 chi, a lowerbreadth of 3chi 6 cun, a depth of 1zhang8 chi and a length of 51824chi. Question: What is the volume?

V.8 Now given a square fort, 1zhang6 chi square and 1zhang5 chi tall.Question: What is the volume ?

V.9 Now given a circular fort with a circumference of 4zhang8 chi and analtitude of 1zhang1 chi. Question: What is the volume ?

Chapter 25

Right Triangles in Zhoubi Suanjing

In the ancient Chinese tradition, a right triangle is called agougu. Gouandgumeanrespectively the shorter and longer legs of a right triangle. The hypotenuse is calledxian.

25.1 The right triangle theorem

From theZhou Dynasty Canon of Gnomonic Computations(Zhoubi suanjing) com-piled around the second century, B.C.:

The circle originates from the square, the square from the rectangle,and the rectangle from9 × 9 = 81. Therefore, cutting up a rectangle,we have (a right triangle) of width 3, length 4, and distance betweentwo corners 5. After erecting a square on this (diagonal), surround itwith (four) half-rectangles to form a square. In this way, wefind (thatthe sides of the right triangle are) 3-4-5: the two rectangles (that makeup the corners) have to increase by 25 (to fill up the whole square). Thisis calledpiling up rectangles.

632 Right Triangles in Zhoubi Suanjing

25.2 A Commentary on the diagram of right triangle,square and circle1

Multiply each of the legs2 to itself, combine them to give the area [of the square]on the hypotenuse.3 The square root is the hypotenuse. According to the diagram,4

the product of the legs can be taken as twice the area of the redtriangle. Doublingit yields four triangles. Multiply the difference of the legs to itself as the area of theyellow square in the middle. Adding this area [from the difference of the legs] tothe four triangles also gives the area [of the square] on the hypotenuse.5

Figure 25.1: The Hypotenuse diagram

Subtract this area [of the square on the difference between the leg] from thearea on the hypotenuse, half the difference, take the ‘square root’ of this, with thedifference [between the legs] as ‘middle coefficient’.6 In this way we get theshorter leg again. To get the longer leg, add the difference to the shorter one.7

Whenever the areas of [the squares on] the shorter and the longer legs are com-bined, they give the area of [the square on] the hypotenuse. Whether as a squareinside, or as a gnomon outside, the shapes may appear singular, yet the quantities8

1gou gu fang yuan tu zhu, a collection of rules for solving problems on right triangles is appendedto theZhoubi Suanjing, attributed to ZHAO Shuang in the third century A.D.

2In the annotations here, we shall usea andb to stand for the legs, andc for the hypotenuse.3This is the main theorem, known as the theorem ofgou gu.4“The diagram for the hypotenuse” (xian tu), associated with the text of Zhoubi, but was lost.

There are various emandations for the diagram. The one we adopt here follows QIAN Baocong.QIAN also associates several supplementary diagrams.

5c2 = 4 · 12ab+ (b− a)2.

6Givenc andb− a, one can finda by solving the equationx(x+ (b− a)) = 12 (c

2 − (b− a)2).But Zhao did not explain how to solve this. It is apparent thatthis process (calleddai cong kai fang),as an extension of the method of extraction of square roots, was presumed to be well known to thereader.

7This paragraph gives thesolutionof a right triangle withc andb− a given.8area.

25.2 2 633

are uniform; the bodies may be different, yet the numbers arethe same.9

a

a c

c

b

b

c

c

c+ bc− b

Gnomon for the shorter leg

The gnomon10 for the area on the shorter leg has as its breadth the difference ofthe hypotenuse and the longer leg, and as its length their sum, and the area for thislonger leg forms the square inside.

Subtracting the area of this gnomon from the area of the square on the hy-potenuse, extracting the square root, we get the longer leg.

Twice the longer leg ‘as middle coefficient’, extracting [square root], the cornerof the gnomon is the difference between the hypotenuse and the longer leg. 11

Adding the longer leg [to this difference] gives the hypotenuse.Dividing the area [of the square] on the shorter leg by the difference [between

the hypotenuse and the longe leg], we get the sum of the hypotenuse and the longerleg. 12

Dividing the area [of the square] on the shorter leg by the sum[of the hypotenuseand the longer leg], we get the difference of the hypotenuse and the longer leg.13

Multiply the sum [of the hypotenuse and the longer leg] by itself. Add thisto the area on the shorter leg. With this sum as dividend, and double the sum[of hypotenuse and the longer leg] as divisor, we get the hypotenuse again.14

Subtracting the area on the shorter leg from the square of thesum [of the hypotenuseand the longer leg], with [this as dividend and] the same divisor [as before], we getthe longer leg.15 16

. . .9The paragraph reiterates the theorem on areas as a basic principle. In what follows the square

on the shorter leg is also construed as the area of a gnomon.10The Chinese word here isju. This is also the same word for “rectangle”. But the context dictates

that the shape is the “carpenter’s square”, hence the translationgnomon. Eucl.II, Definition 2: “Andin any parallelogrammic area let any one whatever of the parallelograms about its diameter with thetwo complements be called a gnomon.”

11Given the longer legb and the areaA of this gnomon, one can find the differencec − b as thesolution ofx(x+ 2b) = A.

12c+ b = a2 ÷ (c− b).13c− b = a2 ÷ (c+ b).14c = [(c+ b)2 + a2]÷ 2(c+ b), becausec = c+b

2 + c−b

2 = c+b

2 + a2

2(c+b) .15b = (c+ b)2 − a2]÷ 2(c+ b), becauseb = c+b

2 − c−b

2 = c+b

2 − a2

2(c+b) .16This paragraph gives the solution of a right triangle with givena and one ofc± b.

634 Right Triangles in Zhoubi Suanjing

Multiply the two differences17 , double it and extract the square root. To theresult add the difference between the hypotenuse and the longer leg, we get theshorter leg.18 Added to the difference between the hypotenuse and the shorter leg,this gives the longer leg.19 Added to the two differences this gives the hypotenuse.20 21

a

a

b

b

c− b

c− a

c− b

c− a

a+ b− c

a+ b− c

From twice the square on the hypotenuse, remove the square onthe differencebetween the longer and the shorter legs. This is the square onthe sum of the legs,as is clear from the [hypotenuse] diagram (xian tu).

Indeed, the double of the area on the hypotenuse fills up the large, outside squarewith a surplus. 22 This surplus is indeed the area on the difference between thelonger and the shorter legs. Subtracting from this [double of the area on the hy-potenuse] the area on the difference, extracting [square root], we get the side of theouter large square.

The side of this large square is the sum of the longer and the shorter legs. Mul-tiply this sum by itself, double, and subtract from the double of the area on thehypotenuse. Extracting [the square root of] the difference, we get the yellow areain the middle. The side of this area is the difference betweenthe longer and theshorter legs.23 Subtract the difference from the sum, half it; we get the shorter leg.Add the difference to the sum, half it; we get the longer leg.24 25

17Between the hypotenuse and each of the two legs.18a =

√

2(c− a)(c− b) + (c− b).19b =

√

2(c− a)(c− b) + (c− a).20c =

√

2(c− a)(c− b) + (c− a) + (c− b).21The reconstruction of the associated diagram is explained in LIU Hui’s commentary onJZSS

9.12.22“the yellow area”.2c2 = (a+ b)2 + (b− a)2.23b− a =

√

2c2 − (a+ b)2.24a = 1

2 (a+ b)− 12 (b− a); b = 1

2 (a+ b) + 12 (b− a).

25The paragraph gives the solution of right triangle withc and one ofb± a given.

25.2 2 635

[Suppose we have a rectangle.] A segment26 is its breadth and width combined.Its area is the short leg and the long leg [of a right triangle]multiplied together.27

Subtract four times the area from the area [of the square on the hypotenuse]. Extract[the square root] of the remainder to get the difference [between the breadth and thewidth]. 28 Subtract this difference from the sum, half the remainder; and we get thebreadth.29 Subtract the breadth from the segment, and we get what we want. 30

26“Twice the hypotenuse” in the text. The Chinese characterbei (for “twice”) apparently is re-dundant. It is inconsistent with the last sentence of the paragraph: “subtract the breadth from thehypotenuse and we get what we want”, namely, the width. The present paragraph presents themethod of breadth and width; it amounts to finding two unknown given their sum and product.Puttingx + y = 2c andxy = a2 for a right triangle with one lega and hypotenuse2c trivializesthe issue here, for then clearly,x − y = 2b andy = c − b, x = c + b. Incidentally, a slight variantof this method leads to the ‘extraction of square root with middle coefficient’: finding two unknownlengths with given product and difference. Perhaps it is forthis reason that the method ofdai congkai fangis never explicitly explained in the ancient Chinese texts.

27To find the breadth, sayy, and the width , sayx, of the rectangle, imagine you have the diagramfor the right triangle with these unknowns lengths as legs.

28x− y =√

(x+ y)2 − 4xy =√l2 − 4ab, wherel is the length of the given segment.

29y = [(x+ y)− (x− y)]÷ 2.30x = l − y.

Chapter 26

Right triangles: Chapter 9 of JZSS

The gougurule and JZSS 9.1-3

JZSS 9.1

Now, given a right triangle, the lengths of itsgouandguare 3chi and 4chi respec-tively.Question: What is the length of the hypotenuse?Answer: 5chi.

JZSS 9.2

Now, given a right triangle, the lengths of its hypotenuse and gouare 5chi and 3chi respectively.Question: What is the length of itsgu?Answer: 4chi.

JZSS 9.3

Now, given a right triangle, the lengths of its hypotenuse and guare 5chi and 4chirespectively.Question: What is the length of itsgou?Answer: 3chi.

The right triangle ( Gougu) L1 rule

Add the squares ofgou andgu, take the square root [of the sum] giving the hy-potenuse.L2 Further, the square ofgu is subtracted from the square on the hy-potenuse. The square root of the remainder is thegou. Further, the square of the

638 Right triangles: Chapter 9 of JZSS

you is subtracted from the square on the hypotenuse. The square root of the remain-der isgu. L3

L1: The shorter side isgou, and the longer sidegu. The side opposite to theright angle is called the hypotenuse (xian). Gouis shorter thangu, andgu is shorterthan the hypotenuse. These apply in all rates (standards), and so are explicated here.

L2: Let the square ongou be red in colour, the square ongu be blue. Letthe deficit and excess parts be mutually substituted into corresponding positions,the other parts remain unchanged. They are combined to form the square on thehypotenuse. Extract the square root to obtain the hypotenuse.

Figure 26.1: LI Huang’s proof of thegougutheorem

L3: The sum of the squares ongouand thegu is the square on the hypotenuse.[If] one of them is missing, the remaining one can be obtainedfrom the rest.

JZSS 9.4

Now given a circular log, 2chi 5 cun in diameter. Assume it is turned into a rectan-gular plank 7cunthick.Question: What is the width?Answer: 2chi 4 cun.Method: Square the diameter 2chi 5 cun, subtract the square of 7cun from it,extract the square root of the remainder. This is the width.L

L: Here, take the diameter 2chi 5 cunas the hypotenuse [of a right triangle], thethickness of the plank 7cunasgou. So its width isgu.

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JZSS 9.5

Now given a tree 2zhanghigh, and 3chi in circumference. A kudzu vine windsaround it 7 times from its root to its top.Question: What is the length of the vine?Answer: 2zhang9 chi.Method: Multiply the circumference by 7 circuits asgu, let the height of the tree begou. Find the hypotenuse as the length of the vine.L

L: Obtain the length of the vine according to the circumference and height of thetree. The vine winds up the tree. Take a blue thread winding round a pen. Straightenthe thread and observe. Every circuit gives a right triangle, with the segment asgu,the circumference asgou and the length of the vine as the hypotenuse. Multiplythe circumference by 7 circuits means combining allgouinto one [large]gou. Heregouis, unexpectedly, longer than thegu, so in the Method we consider reversely theheight of the tree as thegouand the circumference asgu. To find the hypotenusefrom gou andgu also refers to the first figure given under the [ordinary] no tree-circumference condition. It is obvious that the sum of the squares on thegouandgu is the square on the hypotenuse. However, the squares ongu or goumay be putinside the square on the hypotenuse, either in the interior or on the exterior. If put inthe interior, it is a square, if put on the exterior it becomesa gnomon, in these casesthey are different in form but equal in area. In the figure, thered gnomon, whosearea is equal to the square ongou, is put on the exterior, its area is also equal to arectangle with length the sum ofgu and the hypotenuse, and width the differencebetweengu and the hypotenuse. The square on thegu is put in the interior. If theblue gnomon, whose area is equal to the square ongu, is put in the exterior, itsarea is also equal to a rectangle with length the sum ofgouand the hypotenuse, andwidth the difference betweengouand the hypotenuse. The square ongu is put inthe interior. Therefore divide [the square ongouor gu] by the sum or difference [ofgu or gouand the hypotenuse] or multiply the sort [difference] by thelong [sum tofind th unknowngouandgu].


JZSS 9.6

A pond is a 1zhang(= 10 chi) square. A reed grows at its center and extends 1chi out of the water. If the reed is pulled to the side of the pond, it reaches the sideprecisely.Question: What are the depth of the water, and the length of thereed ?Answer: water 12chi deep; reed 13chi long.Method: Square half the side of the pond.L1 From it we subtract the square of 1chi,L2 the height above the water. Divide the remainder by twice theheight abovethe water to obtain the depth of the water.L3 The sum of the result and the heightabove the water is the length of the reed.

L1: Here take half the side of the pond, 5chi, asgou, the depth of the waterasgu, and the length of the reed as the hypotenuse. Obtaingu and the hypotenusefrom gouand the difference betweenguand the hypotenuse. Therefore, squaregoufor the area of the gnomon.

L2: The height above the water is the difference betweenguand the hypotenuse.Subtract the square of this difference from that of the area of the gnomon; take theremainder.

L3: Let the difference between the width of the gnomon and the depth of thewater begu. Therefore construct [a rectangle] with a width of 2chi, twice the heightabove the water. Its length is the depth of water to be found.

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JZSS 9.7 (Hanging rope)

When one end of a rope is tied to the top of a (vertical) pole, 3chi rests on theground. When stretched, the other end of (the rope) is 8chi from the pole.Question: How long is the rope?Answer: 1zhang21

6chi.

Method: Square the distance from the foot,L1 divide by the length of the end lyingon the ground,L2 add the result to the length of the end. Halve it, giving the lengthof the rope.L3

L1: Here consider the distance from the foot, 8chi, asgou, the length of thestring as the hypotenuse. The Problem of “the string tightlystretched” is the sametype as that of “the gate away from the threshold”, i.e. giventhegouand the dif-ference betweengu and the hypotenuse, findgu and the hypotenuse. Square thedistance to get the area of the gnomon.

L2: The end of the string lying on the ground is the difference between thehypotenuse andgu; take it to divide the area of the gnomon to obtain the sum of thehypotenuse and thegu.

L3: If the numerator cannot be halved, double the denominator.The sum addedto the difference is twice the length of the string. So halve the result. Subtract thedifference from the sum and halve it to get the height of the pole.

JZSS 9.8

Now give a wall 1zhanghigh. A pole leans against the wall so that its top is evenwith the top of the wall. If the foot of the pole is moved 1chi further from the wall,the pole will lie flat on the ground.Question: What is the length of the pole?Answer: 5zhang5 cun.Method: Square the height of the wall 10chi and divide it by the distance movedby the foot of the pole; halve the sum to obtain the length of the pole.L


L: Here take 1zhang, the height of the wall, as thegou, the length of the poleas the hypotenuse, and 1chi, the distance moed from the wall as the differencebetweengu and the hypotenuse. The Problem is the same type as that of “a stringhanging from the top of a pole”.

JZSS 9.9

Now given a circular log of unknown size buried in a wall. When sawn 1cundeep,it shows a breadth of 1chi.Question: What is the diameter of the log?Answer: 2chi 6 cun in diameter.Method: Square half of the breadth sawn,L1 divide by its depth. Add the result tothe depth to obtain the diameter.L2

L1: In the Method, take the breadth sawn asgou, the diameter as the hy-potenuse. The depth, 1cun, is half the difference betweengu and the hypotenuse.So the breadth is also halved.

L2: The depth added is also half the difference. In the Problem [7] above oneshould halve the distance to the foot of the pole. Here all thedata have been halved,so the difference is no more halved.

JZSS 9.10

Now given a gate, partially opened, is 1chi away from the threshold. There is a gapof 2 cunbetween the halves [of the gate].Question: What is the width of the gate?Answer: 1zhang1 cun.Method: Square 1chi, the distance from the threshold, divide by half the distancebetween the halves to obtain the width of the gate.L

L: Here, take the distance from the threshold as thegou, half the width of thegate as the hypotenuse, and half the distance between the halves as the differencebetween thegu and the hypotenuse. To find the hypotenuse, the result shouldbehalved. However, the width of the gate is twice the hypotenuse, so it is no longerhalved.

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JZSS 9.11

Now given a door, whose height exceeds the width by 6chi 8 cun. Two [opposite]corners are 1zhangapart.Question: What are the height and the width of the door?Answer: The width 2chi 8 cun; height 9chi 6 cun.Method: Let the square of 1zhangbe shi. Halve the difference; square it, anddouble. Subtract this fromshi. Halve the remainder and extract its square root.Subtract from this half the given difference, giving the thewidth of the door. Add[to this square] half the difference, giving the height of the door.L,

L: Let the width of the door begou, the height begu and the distance betweentwo [opposite] corners be the hypotenuse. The height exceeds the width by 6chi8 cun; this is the difference betweengou and thegu. Calculate according to thediagram. The square of the hypotenuse is exactly 10000 [square] cun. Double it,and subtract from it the square of the difference between thegouandgu. Extract thesquare root of the remainder to obtain the sum of the width andthe height. Subtractthe difference from the sum and halve it to obtain the width ofthe door, to [thewidth] add this difference to obtain the height of the door.

The Method here finds half the dimensions of the door first. Forthere are 4 redsquare areas and 1 yellow square area in a 1zhangsquare. Twice the square ofhalf the difference between the width and height is2

4yellow square area. Half the

difference between theshiand half the yellow square area is 2 red square areas and1

4

yellow square, which is just14

of the big square. So its square root is half the sum ofthe width and height of the door. From that we subtract half the difference betweenthe height and width to obtain the width, and adding that to half the difference weobtain the height of the door.

In the same diagram, multiplygou and gu, double the product, and add thesquare of the difference (ofgou andgu). This also yields the square on the hy-potenuse. [If] the product ofgou and gu is known, then the hypotenuse can bedetermined.1

1This translation follows LI Jimin’s critical edition, which is preferable in view of the subsequent


[A digression on the case of equalgou andgu] If gou andgu are equal, eachmultiplied to itself gives a square, and their sum is the square on the hypotenuse.Added to the double of the square of half of the difference (betweengouandgu).[In this form] the difference still appears. All these multiplied to themselves orto each other as dividend and divisors, and the case of longgu and shortgou, aremerely different streams from the same source. Ifgouandguare each 5, the squareon the hypotenuse is 50. Its square root is 7chi, with remainder 1, not exact. If thehypotenuse is 10, its square is 100, whose half is the square on gouor gu, i.e. 50each. The extraction of its square root is also not exact. That is why one speaks ofcircumference 3 and diameter 1, and square side 5 and diagonal 7. Though theseare not exact, it is still possible to speak of their approximations.

[In the general case, consider] the square of the sum ofgouandgu, and also theproduct ofgouandgu. Subtract from the square 4 times the [latter] area. Extractthe square root. This gives the difference ofgouandgu. Add to the sum and halveto givegu. Subtract the difference from the sum, and halve to givegou. Thegou,gu and hypotenuse are the height, width, and diagonal. All these follow from thisdiagram.

[Take] twice of the hypotenuse as the sum of width and length,and the area ofgougnomon. The width is the difference ofguand hypotenuse.

[Take] the area ofgou gnomon (ju gou zhi mi). With twice of gu ascongfa,extract the square root. This also gives the difference between the hypotenuse andthegu.

Here is one [more] method: subtract the square on the difference ofgouandgufrom the square of the hypotenuse. Halve the remainder [asshi]. With the differenceascongfa, extract the square root. This givesgou.

JZSS 9.12

There are a door of unknown height and width, and a pole of unknown length.Horizontally the pole is 4chi too long, and 2chi vertically. But it can just becarried through the door diagonally.Question: What are the height, width, and the diagonal of the door?Answer: The width is 6chi, height 8ci, and diagonal 1zhang.Method: Double the product of the two excesses [of the bamboopole] and extractits square root. The result plus the excess [of the pole] overthe height is the width.a

digression on the case of equalgou andgu. Here is an alternative translation following standardeditions. In the same diagram, the square of the sum ofgou andgu, added to the square of theirdifference, is twice the square on the hypotenuse. Halve this, and extract the square root to obtainthe hypotenuse. [Furthermore], subtract from twice the square on the hypotenuse the square of thedifference ofgou andgu [and extract square root] to find the sum ofgu andgou. Knowing thehypotenuse [and the difference], it is possible to determine gouandgu.

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The result plus the excess [of the pole] over the width is the height. The result plusboth excesses is the diagonal of the door.L

L: Here consider the width, height and diagonal of the door respectively asgou,gu and hypotenuse of a right triangle. The squares ongouandgu may be put withthe squares on the hypotenuse as an interior square or an exterior gnomon. Observetwo ends of the exterior gnomon outside the square ongou.

The square ongouis [partly] covered by a blue gnomon leaving a yellow square.The area of the two overlapping rectangles between the two gnomons is equal tothat of the yellow square. Each of the rectangles has the difference between thehypotenuse andgou as its length, and the difference between the hypotenuse andguas its width. So twice the product of the two differences is the area of the yellowsquare. Extract its square root to obtain the side. The widthof the blue gnomonoutside is the difference between the hypotenuse and thegu, to which one adds [theside of the yellow square] to obtaingou.

JZSS 9.13

A 1 zhangbamboo breaks and its top reaches ground, 3chi from the bamboo.Question: How tall is the broken bamboo ?Answer:411

20chi.


Method: Square the distance from the baseL1 and divide this by its height.L2 Sub-tract the result from the height and halve the difference to obtain the height of thebreak.L3

L1: Here we consider 3chi, the distance from the base as thegou, the height leftstanding asgu, [and] the length of the fallen tip as the hypotenuse. Findgu usingthegouand the sum ofguand the hypotenuse. Hence first squaregouto obtain thearea of a gnomon [which has the area ofgousquare].

L2: The total height of the bamboo, 1zhang, is the sum ofgu and the hy-potenuse. Divide the square of thegouby that to obtain the difference [betweenguand the hypotenuse].

L3: The Method is the same type as that for “hanging rope”.2 It is an alternativeproblem. Similar to the above Method: square the height, i.e. the sum of the area ofthe square onguplus the hypotenuse. From the square [of the total height] subtractthe square of the distance from the base. Consider the remainder as dividend, twicethe height as divisor. Divide, giving the height of the break.

JZSS 9.14

Now two men stand at the same point. A and B walk with rates 7 and3 respectively.B walks [eastward]. A walks 10bu south, then walks in a northeasterly directiontill the two men meet.Question: What is the distance covered by each?Answer: B walks101

2bueastward and A walks141

2budiagonally to meet him.

Method: Halve the sum of the square of 7 and the square of 3, which gives the rateof A walking diagonally. Subtract this from the square of 7. The remainder is therate of A walking south. The rate of B walking eastwards is 3 times 7.L1 Lay downthe 10bu southwards. Multiply this by the rate of A walking diagonally and againby the rate of B walking east as individual dividends. Divideeach by the rate of Awalking south to obtain both the required distances.L2

L1: Here, take the southward distance asgou, the eastward asgu, and the diag-onal distance asgu and the hypotenuse. The rate forgu is 3. The rate for the sumof gouand the hypotenuse is 7. In order to obtain the rate for the hypotenuse divide

2JZSS 9.7.

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the square onguby the sum ofgouand hypotenuse to obtain the difference betweenthegouand the hypotenuse. Add this to the sum and halve it to get the hypotenuse.Subtract the difference betweengouand the hypotenuse from the hypotenuse, to getgou. Fractions, if any, should be reduced to a common denominator, and simplifiedbefore the required rates are determined. The Rule considered the square on thesum ofgou and the hypotenuse being taken as the red and yellow squares joinedso as to avoid denominators appearing. Turn the blue gnomon with the area ofgusquared into a rectangle, whose length is the sum ofgou and the hypotenuse andwidth is the difference betweengouand the hypotenuse. As a whole, the rectangleis twice the hypotenuse in length, and the sum ofgouand the hypotenuse in width.Cut it horizontally half and half, with one half as the hypotenuse rate. The squareof 7 is considered as the rate for the sum of thegouand the hypotenuse, from whichsubtract the hypotenuse rate and the remainder isgou rate. The point where thetwo men stand is the start of the walk. Both the hypotenuse rateandgou rate are[rectangles] with the sum of the hypotenuse andgouas their length [and with thehypotenuse and thegourespectively as their width]; thengu rate will also be a rect-angle with the sum of the hypotenuse and thegouas its length [and with thegu asits width].

L2: The southward distance, 10bu, is known as the “givengou” ( jian gou). Itshould be multiplied by the hypotenuse rate andgu rate respectively and divided bygourate [to obtain the required hypotenuse andgu].

JZSS 9.15

Now given [a right triagle] whosegouandguare 5buand 12bu respectively.Question: What is the side of the inscribed square?Answer: The side is3 9

17bu.

Method: The product of thegouandgu is the dividend, the sum of thegouandguthe divisor. Divide, giving the side of the square.L

L: The product ofgou by gu comprises three pairs of figures, red, blue andyellow. Put the yellow figures at the bottom, and combine the red and blue figuresas rectangles at the top, with the side of the central yellow square as the width, andthe sum ofgouandgu as the length. That is why one adds the sum ofgouand the


gu as divisor. In the figure, the square is contained in the givenright triangle. Tothe top and to the right of the square there appear respectivesmaller right triangles.The relations between their sides retain the same rates as inthe original triangle.The sum of thegouandgu of the small triangle [blue] ongu [side] is equal to thegu. Let guof the original triangle be the sought rate, and the givengou, 5 bu, be thegiven number. Apply the Rule of Three, giving the side of the inscribed square.

Alternatively, letgouof the original triangle be the sought rate, the sum ofgouandgu the given rate, andgu, 12 bu, the given number. Apply the Rule of Threegiving the side of the inscribed square. Although the statement lacks derivation,one arrives at the Method using the divisor and dividend. In the next Problem aboutan inscribed circle, one can see how this is done using the Ruleof Three and theProportional Distribution Rule.

JZSS 9.16

Now given [a right triangle] whosegouandguare 8buand 15bu respectively.Question: What is the diameter of the inscribed circle?Answer: The side is6 bu.Method: 8bu is thegou, 15 bu is gu. Find the hypotenuse. Add these three asdivisor. Takegou to multiply thegu. Take twice [the product] as dividend. Divide,giving the diameter.L

L: The product ofgouby gu is the chief subject in the figure, in which there arethree pairs of figures, red, blue and yellow. Doubling them, get four of each. Copythem onto a small piece of paper, and cut them out. Arrange them upright, slant orupside down by attaching the equal sides together so as to form a rectangle with thediameter as the width and the sum ofgou, guand the hypotenuse as the length. Thatis why we addgou, go and hypotenuse as divisor. From the figure, [the tip of] theblue triangle on thegu is equidistant fromgou, gu and the hypotenuse. Measuring

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this distance ongou andgu by a compass horizontally and vertically, one surelygets a small square.

gou gu hypotenuse

diameter

Further, draw a “middle hypotenuse” [parallel to the hypotenuse] with a com-pass.3 In the middle of [each of]gou andgu there are small right triangles. The

3The recent critical edition of LI Jimin hasyi gui tu hui, though DAI Zhen’s edition hadyi gui qihui.


smallergu [of the right triangle] ongou [of the original triangle] andgou [of theright triangle] ongu [of the given triangle] are both sides of the small square, i.e.half the diameter of the [inscribed] circle. Therefore the [relevant] data are all pro-portional. Let thegou, gu and the hypotenuse be the rates (lie cui), and take theirsum as divisor.

Multiply gou by the various summands as dividend, and divide. Thegu ofthe right triangle ongou [of the given triangle] can be found. Multiplygu by thecorresponding rate as dividend.4 This gives thegou[of the right triangle] ongu [ofthe given triangle, as the product ofgouandgu, divided by the sum ofgou, gu, andhypotenuse. The diameter is twice of this]. Though differently expressed, the waysto form the dividend and divisor are actually the same.

It is also possible to make use of the result of the previous problem on the inscribed square.

Consider the right triangle with the “middle hypotenuse”. Its perimeter is the sum ofgou andgu.

The side of its inscribed square is therefore the product of the side of the inscribed square of the

given triangle and the sum ofgouandgu, divided by the sum ofgou, guand hypotenuse. This is the

product ofgouandgu, divided by the sum ofgou, gu, and hypotenuse. Doubling this, we obtain the

diameter of the circle inscribed in the given right triangle.

Alternatively, subtract the difference betweengu and the hypotenuse from thegou; subtract the difference betweengouand the hypotenuse fromgu; or subtractthe hypotenuse from the sum of thegouandgu, to obtain the required diameter ofthe inscribed circle.

Double the product of the difference betweengou and the hypotenuse by the

4This means that the sum ofgou, gu, hypotenuse of the right triangle ongou is equal togouofthe given triangle; similarly forgu. This is clear by considering the bisectors of the acute angles ofthe given right triangle.

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difference betweengu and the hypotenuse. Extract the square root, giving the di-ameter.

This is JZSS 9.12.

JZSS 9.17

Now a given city 200bu square, with gates opening in the middle of each side. 15bu from the east gate there is a tree.Question: At how manybu from the south gate with one see the tree?Answer:6662

3bu.

Method: The number ofbu from the east gate is the divisor.L1 Square half the sideof the city as dividend. Divide giving the number ofbu. L2

tree

L1: Thegourate is the divisor.L2: Here take 15bu, the distance from the east gate, asgou rate; 100bu, the

distance between the east gate and the southeast corner, asgu rate; and 100bu, thedistance between the south gate and the southeast corner as the givengouFrom thelast, findgu as the number ofbu from the south gate. The square of half the side isequivalent to the product ofgu rate by the givengou. The two are equal.


JZSS 9.18

Now given a city which is 7li from east two west, 9li from south to north, withgates opening in the middle of each side. There is a tree 15li from the east gate.Question: At how manybu from the south gate with one see the tree?Answer:315 bu. 5

Method: Take the distance between the east gate and the [southeast] corner to mul-tiply the distance between the south gate and the [southeast] corner as dividend.The distance between the tree and the [east] gate as divisor.Divide, giving thedistance.L

tree

L: Here take the distance between the east gate and the [southeast] corner, 4li ,asgourate, the distance from the east gate, 15li , asgu rate. The distance betweenthe south gate and the [southeast] corner, 3li , as the givengu. From these findgou;i.e. the required number ofbu from the south gate. The Method is the same as inthe previous problem.

JZSS 9.19

Now given a square city to be measured. It has gates opening inthe middle of eachside. 30bu from the north gate there is a tree. At 750bu from the west gate one cansee the tree.Question: What is the length of each side?Answer:1 li .Method: Quadruple the product (shi) of the two distances from the gates. Extractthe square root of theshi to obtain the length of each side.L

L: According to the previous Method. square half the side; divide it by thedistance from the east gate to obtain the distance from the south gate. Here the

51 li = 300bu

653

tree

product of the two distances is the square of half the side, i.e. one-fourth of the areaof the city. Therefore quadruple it to obtain the whole area.Extract its square rootto obtain the side.

JZSS 9.20

Now given a square city of unknown side, with gates opening inthe middle. 20bufrom the north gate there is a tree, which is visible when one goes 14bu from thesouth gate and then 1775buwestward.Question: What is the length of each side?Answer:250 bu.Method: Take the distance from the north gate to multiply thewestward distance.Double the product as theshi. L1 Take the sum of the distance from the north andsouth gates as the linear coefficient (congfa). Extract the root to obtain the side ofthe city.L2

tree

L1: Here take the westward distance [1775bu] as thegu; the distance from thetree to the south gate, 14bu, as thegou. Take the distance from the north gate, 20bu, asgourate; the distance from the north gate to the western corner asgu rate; i.e.half the side of the city. Then take thegou rate, the distance from the north gate,to multiply thegu, the distance west. The product is the area of a rectangle whosewidth and length aregu rate, half the side of the city, andgou respectively. It is,however, only the area of the western half of the city. Doubling it means adding theeastern half to it for the whole.

L2: In the Method the width from east to west of the rectangle is the same asthe side of the city; the length from south to north is the distance from the tree upto 14bu from the south gate. The two [small] rectangles have the sum of the twonorth and south distances as width and the city side as length. Then the sum of


the two north-south distances [i.e. the difference betweenthe length and width] isconsidered as the linear coefficient. The area is the surplusoutside the city square.

JZSS 9.21

A city is a 10li 6 square. At the center of each side is a gate. A and B part from thecenter of the city. B walks eastwards and A begins southwardsfor some distanceand then turns (in some northeast direction) through a corner of the city to meet B.Their speeds are in the ratio 5:3. How long has each walked ?Answer: A walks800 bu from the south gate and then northeastward48871

2buuntil

he meets B. B walks43121

2bueastwards.

Method: Let 5 be squared and 3 squared also, halve their sum toget the rateof the northeast advance. Subtract this rate from 5 squared.The remainder is therate of the southward advance. The product of 3 by 5 is the rateof B walkingeastward.L1 Lay down half the side, multiply [it] by the rate of the southwardadvance; divide it by the rate of the eastward advance to obtain the number ofbufrom the south gate.L2 Add this to the half side to obtain the number ofbusouthwardfor A. L3 Lay down the number ofbusouthward; to find the hypotenuse, multiply bythe rates of the northeastward; to find B multiply the rate of the eastward advance.Take the products as dividends. Respectively divide by the rate of the southwardadvance to obtain the required numbers ofbu. L4

L1: The Method to find the three rates is the same as in the above Problem(4)in finding the number ofbu for A and B.

L2: 5 li , the distance from the south gate to the southeast corner, i.e. half theside of the city, is considered a smallgu. Required is the number ofbu from the

61 li = 300bu.

655

south gate, so multiply half the side of the city bygourate of the southward advanceand divide it by thegu rate.

L3: To take half the side means starting from the centre of the city.L4: This Method is the same as that of the previous Problem in finding the

number ofbu for A and B.

JZSS 9.22

Now a tree is an unknown distance away from a person. Stand four poles, 1zhangapart, making the left two aligned with the observed object.Observe from the right-rear pole, [the line of sight] cuts 3cun into the left of the front-right pole.Question: How far away is the tree from the person?Answer:33 zhang3 chi 31

3cun.

Method: Consider 1zhangsquared as dividend, 3cunas divisor; and divide.L

L: Here take 3cun, the length cut by the line of sight into the left of the front-right pole, asgou rate; the distance between the two right poles asgu rate; thedistance between the left and right poles as the given you. The distance betweenthe tree and the person is thegu corresponding to the givengou. The gu rate ismultiplied by the givengou, both of which are 1zhang, so the Method says square.Divide by 3cunto obtain the number ofcun.


JZSS 9.23

Now to the west of a tree there is a hill whose height is unknown. The distancebetween the tree and the hill is 53li and the tree is 9zhang5 chi high. A personstanding 3li to the east of the tree observes that the summit of the hill andthe tree-top are aligned. Assume his eyes are at the height 7chi.Question: What is the height of the hill?Answer: 164zhang9 chi 62

3cun.

Method: Lay down the height of the tree [95chi]; subtract the height of the eyes7 chi. Take the remainder to multiply 53li as dividend. Take the distance, 3li ,from the person to the tree as divisor. Divide, and add the height of the tree to thequotient to obtain the height of the hill.L

L: By the gouguRule. from the height of the tree subtract that of the eyes.The remainder 8zhang8 chi is considered asgou rate, 3li , the distance from thesurveyor to the tree asgu rate; and 53li , the distance from the tree to the hill, as thegivengu. Findgou, to which add the height of the tree to get the height of the hill.

JZSS 9.24

Now given a well 5chi in diameter with unknown depth. Stand a pole of 5chi highon the mouth of the well. When one looks down from the tip of the pole to the edgeof the water, the line of sight cuts the diameter 4cun.Question: What is the depth of the well?Answer: 5zhang7 chi 5 cun.Method: Lay down 5chi, the diameter of the well; subtract 4cun, the length cut by

657

the line of sight. Take the remainder to multiply 5chi; take the height of the poleas dividend; 4cun, the length cut by the line of sight, as divisor. Divide, giving thedepth of the well.L

L: Here take the length cut by the line of sight as thegourate, the height of thepole asgu rate, and 4chi 6 cun, the remainder, as the given you. The required depthis gucorresponding to the givengou.

Chapter 27

Haidao Suanjing: LIU Hui’sAppendix to JZSS

Commentary on the sun altitude diagram in ZhoubiSuanjing

sun

A

D

C

B

sun

The words of Zhao Junqing. Yellow A and Yellow B, their areas are equal. Mul-tiply the height of the pole and the distance between the two poles. [The product] isthe area of Yellow A. With the difference of the shadows as thewidth of [a rectanglewith the same area as] Yellow A, and divide. The result is the length of [a rectanglewith the same area as] Yellow A. The top of the rectangle has the same altitude ofthe sun. According to the diagram, the height of the pole should be added. Now[the altitude of the sun] is said to be 80000li ; it is reckoned by adding the length ofthe pole to what is above. Blue C and Blue D, their areas are also equal. Yellow Aand Blue C are adjacent; Yellow B and Blue D are [also adjacent].Their areas areequal. Both have the difference of the shadows as width.

660 Haidao Suanjing: LIU Hui’s Appendix to JZSS

HDSJ 1: The sea island problem

Now to view a sea island [AB], erect two poles [DE, FG] of the same height 3zhang, and 1000bu apart. Assume that the rear pole is aligned with the front pole.Move 123bu backward from the front pole [DE] and view the peak [A] of the is-land from ground level [atH]. It is seen that the tip [D] of the front pole coincideswith the peak [A]. Move 127bu backward from the rear pole [FG] to view [fromI] at ground the peak [A] of the island, which again coincides with the tip [F ] ofthe pole.Question: What are the height [AB] of the island, and how far is it from the pole[DE]? 1

B

A

E

D

G

F

I B

A

E

D

G

F

IH

JK

Method: With the height [EG] of the pole as dividend, the difference [GI −EH] as divisor, divide. Add the height to the sum to get the height of the island. Tofind the distance from the island to the front pole, multiply the backward movementfrom the front pole and separation of the poles as dividend, the difference [GI −EH] as divisor, divide. This gives the number ofli from the island to the [front]pole.

AB =EG×DE

GI − EH+DE,

BE =EG× EH

GI − BH.

Proof. Note that (1) rect.FJ = rect.FB, and (2) rect.DK = rect.DB. Subtrac-tion gives

rect.FJ − rect.DK = rect.EF,

or

GI × (AB −DE)− EH × (AB −DE) = EG×DE,

AB −DE =EG×DE

GI − EH.

1Answer: The height of the island is 4li 55bu. It is 102li 150bu from the pole.

661

HDSJ 2: The pine tree problem

Now there is a pine tree [AB] of unknown height on a hill [BC]. Erect two poles[DE andFG] of the same height 2zhang, front and rear at a distance [EG] 50 buapart. Align the rear pole [FG] with the front pole [DE]. From the front pole movebackward 7bu 4 chi, and view the pine treetop from [H] at the ground level. It isobserved that the treetop [A] coincides with the tip [D] of the pole. Also, view thefoot [B] of the pine tree; it is seen to be at a point [J so thatDJ ] 2 chi 8 cun fromthe tip [D] of the pole. Again, from the rear poleFG] move backward 8bu 5 chito view the pine treetop [A] from [I] at the ground level. It is noted that the pinetreetop also coincides with the tip [F ] of the pole.Question: What are the height [AB] of the pine tree and the distance [CE] fromthe hill to the [front pole]?2

A

B

C

D

E

F

G IH

J

A

B

C

D

E

K

L

N

H

J

M

Method:

AB =EG×DJ

GI − EH+DJ,

CE =EG× EH

GI − EH.

Proof. Compare with the seaisland diagram. RegardingAC as the seaisland, weeasily obtainCE. In the supplementary diagram, rect.DK = rect. DC and rect.JL = rect.JC. From these, rect.MK − rect.JN = rect.JO, and

EH × AB − EH ×DJ = CE ×DJ,

AB =CE ×DJ

EH+DJ

=EG×DJ

GI − EH+DJ.

2Answer: The height of the pine is 12zhang2 chi 8 cun. The hill is 1li 28 47 bu from the pole.

662 Haidao Suanjing: LIU Hui’s Appendix to JZSS

HDSJ 3: Viewing a distant city

Now in the south there is a square city [ABCD] of unknown size. Erect at the eyelevel two poles [E andF ] 6 zhangapart in the east-west direction and join themby a string. Align the east pole [F ] with the southeast and northeast corners [CandB] of the city. From the east pole move northward 5bu [to G] and view onthe northwest corner [A] of the city; the line of observation intersects the string ata point [I] 2 zhang2 chi 61

2cun from its east end. Again, move northward 13bu

2 chi from the pole and view [fromH] the northwest corner of the city. It is foundthat the corner coincides with the west pole [E].Question: What is the [length of the] side of the square city, and how far is the cityfrom the pole?3

FE

G

I

D

A

C

B

H

K

J

Method:

KH =IF × FH

EF,

AB =IF ×GH

KH −GF,

BF =FG×GH

KH −GF.

3Answer: The side of the squared city measures 31i 43 34 bu. The city lies 41i 45 bu from the

poles.

MST Topics in History of Mathematicsmath.fau.edu/yiu/MSTHM2017/17HM726B.pdf602 The ancient Chinese...

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